Simplify; express your answer in exponential form. Assume $k\neq 0, q\neq 0$. $\dfrac{{k^{4}q^{-1}}}{{(k^{-3}q^{-5})^{2}}}$
To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${k^{4}q^{-1} = k^{4}q^{-1}}$ On the left, we have ${k^{4}}$ to the exponent ${1}$ . Now ${4 \times 1 = 4}$ , so ${k^{4} = k^{4}}$ Apply the ideas above to simplify the equation. $\dfrac{{k^{4}q^{-1}}}{{(k^{-3}q^{-5})^{2}}} = \dfrac{{k^{4}q^{-1}}}{{k^{-6}q^{-10}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{4}q^{-1}}}{{k^{-6}q^{-10}}} = \dfrac{{k^{4}}}{{k^{-6}}} \cdot \dfrac{{q^{-1}}}{{q^{-10}}} = k^{{4} - {(-6)}} \cdot q^{{-1} - {(-10)}} = k^{10}q^{9}$